A can contains a mixture of two liquids, A and B, in the ratio 7 : 5. When 9…

Concept: When a homogeneous mixture is partially removed, each liquid in it is removed in the same ratio as it exists in the mixture at that moment. If the removed volume is then replaced by only one of the two liquids, that liquid's quantity increases while the other's decreases only due to the removal — so track each liquid's quantity separately through both the removal and the replacement, then use the final given ratio to set up and solve an equation for the unknown.

1. Let the initial quantities of A and B be 7x and 5x litres respectively, since A : B = 7 : 5. Total initial mixture = 12x litres.

2. When 9 litres of mixture is drawn off, it is removed in the same 7 : 5 ratio, so: A removed = 9 x 7/12 = 21/4 litres, and B removed = 9 x 5/12 = 15/4 litres.

3. Quantities remaining after removal: A = 7x - 21/4 litres; B = 5x - 15/4 litres.

4. The can is then filled with 9 litres of liquid B only, so B becomes (5x - 15/4) + 9 = 5x + 21/4 litres, while A stays at 7x - 21/4 litres.

5. Using the new ratio A : B = 7 : 9, form the equation: (7x - 21/4) / (5x + 21/4) = 7/9.

6. Cross-multiplying: 9(7x - 21/4) = 7(5x + 21/4) => 63x - 189/4 = 35x + 147/4 => 28x = 336/4 = 84 => x = 3.

7. So the initial quantity of liquid A = 7x = 7 x 3 = 21 litres.

Verify: initial A = 21 L, B = 15 L (total 36 L). Removing 9 L of mixture in the 7 : 5 ratio takes out 5.25 L of A and 3.75 L of B, leaving A = 15.75 L and B = 11.25 L. Adding 9 L of B gives B = 20.25 L. The ratio 15.75 : 20.25 simplifies (dividing both by 2.25) to 7 : 9 - matching the given condition, confirming the initial quantity of A is 21 litres.